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3 years ago

The line L1 has equation 2x+y=8.The line L2 passes through the point A(7,4) and is perpendicular to L1.

Mathematics

1 answer:

baherus [9]3 years ago

4 0

If m1 and m2 are the slopes of each line, for them to be perpendicular you gotta solve
m1 * m2 = -1

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A teacher gave her class two exams; 60% of the class passed the second exam, but only 48% of the class passed both exams. What p

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Match each expression with the equivalent expanded expression.

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Question

Match each expression with the equivalent expanded expression.

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Step-by-step explanation:

Step-by-step explanation:

Question

Match each expression with the equivalent expanded expression.

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Step-by-step explanation:

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2 years ago

The maximum value of the function: f(x)= -5 x ^2 +30x-200 is?

VARVARA [1.3K]

Answer:

$\displaystyle - 155$

Step-by-step explanation:

we are given a quadratic function

$\displaystyle f(x) = - 5 {x}^{2} + 30x - 200$

we want to figure out the minimum value of the function

to do so we need to figure out the minimum value of x in the case we can consider the following formula:

$\displaystyle x _{ \rm min} = \frac{ - b}{2a}$

the given function is in the standard form i.e

$\displaystyle f(x) = a {x}^{2} + bx + c$

so we acquire:

b=30
a=-5

thus substitute:

$\displaystyle x _{ \rm min} = \frac{ - 30}{2. - 5}$

simplify multiplication:

$\displaystyle x _{ \rm min} = \frac{ - 30}{ - 10}$

simply division:

$\displaystyle x _{ \rm min} = 3$

plug in the value of minimum x to the given function:

$\displaystyle f (3)= - 5 {(3)}^{2} + 30.3 - 200$

simplify square:

$\displaystyle f (3)= - 5 {(9)}^{} + 30.3 - 200$

simplify multiplication:

$\displaystyle f (3)= - 45 + 90- 200$

simplify:

$\displaystyle f (3)= - 155$

hence,

the minimum value of the function is -155

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2 years ago